A body cools from `70^@C "to" 40^@C` in 5 min. Calculate the time it takes to cool from `60^@C "to" 30^@`C . The temperature of the surroundings is `20 ^@C`.

Options

A. 1 min
B. 7 min
C. 6 min
D. 15 min

Answer

B

Explanation

Body cools from `70^@ C "to " 40^@ C ` in 5 minute, hence, `T_1 = 70 ^@ C,T_2 = 40^@ C and t= 5 ` minute
Temperature of surrounding , `T_0 = 20 ^@C`
By Newton's law of cooling,
`(T_1-T_2)/t = k [ (T_1+T_2)/2 -T_0]`
or, `(70- 40)/5 =k [ (70+40)/2 -20] rArr k = 6/(35)` ......(i)
Again, let body cools from ` 60^@C "to" 30^@C` in time t minutes.
i,c , `T_1^' = 60^@C and T_2^' = 30^@C`
`therefore` By Newton's law of cooling ,
`(T_1^'-T_2^')/t-k [(T_1^'+T_2^')/2-T_0]`
`(60-30)/t=6/(35)[(60+30)/2-20] [because` From Eq. (i)]
`(30)/t =6/(35)[25] rArr t =7` minutes